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\vskip 1cm{\LARGE\bf
Efficient Lower Bounds on the Number \\
\vskip .1in
of Repetition-free Words
}
\vskip 1cm
{\large
Roman Kolpakov\footnote{This work is supported
by the program of the President of the Russian Federation
for supporting of young researchers and scientific
schools (Grants MD--3635.2005.1 and NSh--5400.2006.1)
and the Russian Foundation for Fundamental Research
(Grant 05--01--00994).}\\
Lomonosov Moscow State University \\
Vorobjovy Gory \\
119992 Moscow \\
Russia \\
\href{mailto:foroman@mail.ru}{\tt foroman@mail.ru}
}
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\vskip 1cm
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{lemma}{Lemma}[section]
\begin{abstract}
We propose a new effective method for obtaining lower bounds on the
number of repetition-free words over a finite alphabet.
\end{abstract}
%\newtheorem{proposition}{Proposition}
\section{Introduction}
Repetition-free words over finite alphabets are a traditional object of
research in combinatorics of words. A word~$w$ over an alphabet~$\Sigma$
is a finite sequence $a_1\cdots a_n$ of symbols from~$\Sigma$. The number~$n$
is called the {\it length} of~$w$ and is denoted by $|w|$. The symbol
$a_i$ of~$w$ is denoted by $w[i]$. A word $a_i\cdots a_j$, where $1\le
i\le j\le n$, is called a {\it factor} of~$w$ and is denoted by $w[i:j]$.
For any $i=1,\ldots,n$ the factor $w[1:i]$ ($w[i:n]$) is called a {\it prefix}
(a {\it suffix}) of~$w$. A positive integer $p$ is called a {\it period} of~$w$
if $a_i=a_{i+p}$ for each $i=1,\ldots ,n-p$. If $p$ is the minimal period
of~$w$, the ratio $n/p$ is called the {\it exponent} of~$w$. Two words
$w',w''$ over~$\Sigma$ are called {\it isomorphic} if $|w'|=|w''|$ and
there exists a bijection $\sigma:\Sigma\longrightarrow\Sigma$ such that
$w''[i]=\sigma (w'[i])$, $i=1,\ldots,|w'|$. The set of all words
over~$\Sigma$ is denoted by $\Sigma^*$.
For any finite set~$A$ we denote by $|A|$ the number of elements of~$A$.
Let $W$ be an arbitrary set of words. This set is called {\it factorial}
if for any word~$w$ from~$W$ all factors of~$w$ are also contained in~$W$.
We denote by $W(n)$ the subset of~$W$ consisting of all words of length~$n$.
If $W$ is factorial then it is not difficult to show (see, e.g.,~\cite{Brink,
BaElGr}) that there exists the limit $\lim_{n\to\infty} \sqrt[n]{|W(n)|}$
which is called the {\it growth rate} of words from~$W$. For any word~$v$
and any $n\ge |v|$ we denote by $W^{(v)}(n)$ the set of all words from $W(n)$
which contain~$v$ as a suffix.
By a {\it repetition} we mean any word of exponent greater than~1.
The best known example of a repetition is a {\it square}; that is,
a word of the form $uu$, where $u$ is an arbitrary nonempty word.
Avoiding ambiguity\footnote{Note that the period of a square is not
necessarily the minimal period of this word.}, by the {\it period}
of the square $uu$ we mean the length of~$u$. In an analogous way,
a {\it cube} is a word of the form $uuu$ for a nonempty word~$u$,
a the {\it period} of this cube is also the length of~$u$. A word
is called {\it square-free} ({\it cube-free}) if it contains no squares
(cubes) as factors. It is easy to see that there are no binary square-free
words of length more than~3. On the other hand, by the classical results
of Thue~\cite{Thue06,Thue12}, there exist ternary square-free words of
arbitrary length and binary cube-free words of arbitrary length. If we
denote by $S^{\langle\mbox{sf}\rangle}(n)$ the number of ternary square-free
words of length~$n$ and by $S^{\langle\mbox{cf}\rangle}(n)$ the number
of binary cube-free words of length~$n$, we then have that
$S^{\langle\mbox{sf}\rangle}(n)>0$ and $S^{\langle\mbox{cf}\rangle}(n)>0$
for any~$n$. For ternary square-free words this result was strengthened
by Dejean in~\cite{Dejan}. She found ternary words of arbitrary length
which have no factors with exponents greater than $7/4$. On the other hand,
she showed that any long enough ternary word contains a factor with an
exponent greater than or equal to $7/4$. Thus, the number $7/4$ is the
minimal limit for exponents of prohibited factors in arbitrarily long
ternary words. For this reason we call ternary words having no factors
with exponents greater than $7/4$ {\it minimally repetitive} ternary
words. Dejean conjectured also that the minimal limit for exponents of
prohibited factors in arbitrarily long words over a $k$-letter alphabet
is equal to $7/5$ for $k=4$ and $k/k-1$ for $k\ge 5$. This conjecture
was proved for $k=4$ by Pansiot~\cite{Pans}, for $5\le k\le 11$
by Moulin Ollagnier~\cite{Olag}, for $12\le k\le 14$ by Mohammad-Noori
and Currie~\cite{Moh-Noor}, and for $k\ge 38$ by Carpi~\cite{Carpi}.
Denote by $S^{\langle\mbox{lf}\rangle}(n)$ the number of minimally repetitive
ternary words of length~$n$. It follows from the result of Dejean that
$S^{\langle\mbox{lf}\rangle}(n)>0$ for any~$n$.
Note that the set of all ternary square-free words, the set of all
binary cube-free words, and the set of all minimally repetitive
ternary words are factorial. So there exist the growth rates
$\gamma^{\langle\mbox{sf}\rangle}=\lim_{n\to\infty}
\sqrt[n]{S^{\langle\mbox{sf}\rangle}(n)}$,
$\gamma^{\langle\mbox{cf}\rangle}=\lim_{n\to\infty}
\sqrt[n]{S^{\langle\mbox{cf}\rangle}(n)}$,
$\gamma^{\langle\mbox{lf}\rangle}=\lim_{n\to\infty}
\sqrt[n]{S^{\langle\mbox{lf}\rangle}(n)}$ of words from these sets.
Brandenburg proved in~\cite{Brand} that the values
$S^{\langle\mbox{sf}\rangle}(n)$ and $S^{\langle\mbox{cf}\rangle}(n)$
grew exponentially with~$n$, namely, $S^{\langle\mbox{sf}\rangle}(n)\ge
6\cdot 1.032^n$ and $S^{\langle\mbox{cf}\rangle}(n)\ge 2\cdot 1.080^n$,
i.~e. $\gamma^{\langle\mbox{sf}\rangle}\ge 1.032$ and $\gamma^{\langle
\mbox{cf}\rangle}\ge 1.080$. Later the lower bound for $\gamma^{\langle
\mbox{sf}\rangle}$ was improved consecutively\footnote{A review of
results on the estimations for the number of repetition-free words
can be found in~\cite{Berst}.} by Ekhad, Zeilberger, Grimm, and Sun
in~\cite{EkhZeil,Grimm,Sun}. The best upper bounds known at present
$\gamma^{\langle\mbox{sf}\rangle}<1.30178858$ and
$\gamma^{\langle\mbox{cf}\rangle}<1.4576$ are obtained by Ochem and
Edlin in \cite{OchemWOWA} and~\cite{Edlin} respectively. In~\cite{OchemTIA}
Ochem established the exponential growth of the number of minimally
repetitive words over either a three-letter or a four-letter
alphabet. However, this result does not give any significant lower
bound for $\gamma^{\langle\mbox{lf}\rangle}$.
In~\cite{WOWA} we proposed a new method for obtaining lower bounds on
the number of repetition-free words. This method is essentially based on
inductive estimation of the number of words which contain repetitions as
factors. Using this method, we obtained the bounds $\gamma^{\langle\mbox{sf}
\rangle}\ge 1.30125$ and $\gamma^{\langle\mbox{cf}\rangle}\ge 1.456975$.
The main drawback of the proposed method was the large size of computer
computations required. In particular, for this reason we did not
manage to obtain
a lower bound for $\gamma^{\langle\mbox{lf}\rangle}$ by the proposed method.
In this paper we propose a modification of the given method which requires
a much less size of computer computations. Using this modification,
we obtain the bounds $\gamma^{\langle\mbox{sf}\rangle}\ge 1.30173$,
$\gamma^{\langle\mbox{lf}\rangle}\ge 1.245$, and $\gamma^{\langle\mbox{cf}
\rangle}\ge 1.457567$. Comparing the obtained lower bounds for $\gamma^{\langle
\mbox{sf}\rangle}$ and $\gamma^{\langle\mbox{cf}\rangle}$ with the known upper
bounds, one can conclude that we have estimated $\gamma^{\langle\mbox{sf}\rangle}$
and $\gamma^{\langle\mbox{cf}\rangle}$ within a precision of $0.0001$, which
demonstrates the high efficiency of the proposed modification.
\section{Estimation for the number of ternary square-free words}
For obtaining a lower bound on $\gamma^{\langle\mbox{sf}\rangle}$ we
consider the alphabet $\Sigma_3=\{0,1,2\}$. Denote the set of all
square-free words from~$\Sigma^*_3$ by~${\cal F}$. Let $m$ be a
natural number, $m>2$, and $w',w''$ be two words from ${\cal F}(m)$.
We call the word $w''$ a {\it descendant} of the word $w'$ if
$w'[2:m]=w''[1:m-1]$ and $w'w''[m]=w'[1]w''\in {\cal F}(m+1)$.
The word $w'$ is called in this case an {\it ancestor} of the
word~$w''$. Further, we introduce a notion of closed words in
the following inductive way. A word $w$ from ${\cal F}(m)$ is
called {\it right closed} ({\it left closed}) if and only if
this word satisfies one of the two following conditions:
(a) {\bf Basis of induction.} $w$ has no descendants (ancestors);
(b) {\bf Inductive step.} All descendants (ancestors) of~$w$
are right closed (left closed).
\noindent
A word is {\it closed} if it is either right closed or left closed.
Denote by ${\cal L}_m$ the set of all words from $\Sigma^*_3$
which do not contain closed words from ${\cal F}(m)$ as factors.
By ${\cal F}_m$ we denote the set of all square-free words from
${\cal L}_m$. Note that a word $w$ is closed if and only if any
word isomorphic to~$w$ is also closed. So we have the following
obvious fact.
\begin{proposition}
For any isomorphic words $w',w''$ and any $n\ge |w'|$ the
equality $|{\cal F}_m^{(w')}(n)|=|{\cal F}_m^{(w'')}(n)|$
holds.
\label{utvizo}
\end{proposition}
\noindent
By ${\cal F}'(m)$ we denote the set of all words~$w$ from ${\cal F}(m)$
such that $w[1]=0$ and $w[2]=1$. It is obvious that for any word~$w$
from ${\cal F}(m)$ there exists a single word from ${\cal F}'(m)$ which
is isomorphic to~$w$. Let $w',w''$ be two words from ${\cal F}'(m)$.
We call the word $w''$ a {\it quasi-descendant} of the word $w'$ if
$w''$ is isomorphic to some descendant of~$w'$. The word $w'$ is
called in this case a {\it quasi-ancestor} of the word~$w''$. Let
${\cal F}''(m)$ be the set of all words from ${\cal F}'(m)$
which are not closed. Since ternary square-free words of arbitrary
length exist, ${\cal F}''(m)$ is not empty for any~$m$. Denote
$s=|{\cal F}''(m)|$. We enumerate all words from ${\cal F}''(m)$
by numbers $1, 2,\ldots,s$ in lexicographic order and denote
$i$-th word of the set ${\cal F}''(m)$ by $w_i$, $i=1,\ldots ,s$.
Then we define a matrix $\Delta_m=(\delta_{ij})$ of size $s\times s$
in the following way: $\delta_{ij}=1$ if and only if $w_i$ is an
quasi-ancestor of $w_j$; otherwise $\delta_{ij}=0$. Note that $\Delta_m$
is a nonnegative matrix, so, by Perron-Frobenius theorem, for
$\Delta_m$ there exists some maximal in modulus eigenvalue~$r$ which
is a nonnegative real number. Moreover, we can find some eigenvector
$\tilde x=(x_1;\ldots; x_s)$ with nonnegative components which
corresponds to~$r$. Assume that $r>1$ and all components of~$\tilde x$
are positive. For $n\ge m$ define $S^{\langle\mbox{sf}\rangle}_m(n)=
\sum_{i=1}^s x_i\cdot |{\cal F}_m^{(w_i)}(n)|$. In an inductive way
we estimate $S^{\langle\mbox{sf}\rangle}_m(n+1)$ by
$S^{\langle\mbox{sf}\rangle}_m(n)$.
First we estimate $|{\cal F}_m^{(w_i)}(n+1)|$ for $i=1,\ldots ,s$.
It is obvious that
\begin{equation}
|{\cal F}_m^{(w_i)}(n+1)|=|{\cal G}^{(w_i)}(n+1)|-|{\cal H}^{(w_i)}(n+1)|,
\label{Fwin}
\end{equation}
where ${\cal G}^{(w_i)}(n+1)$ is the set of all words~$w$ from
${\cal L}_m^{(w_i)}(n+1)$ such that the words $w[1:n]$ and $w[n-m+1:n+1]$
are square-free, and ${\cal H}^{(w_i)}(n+1)$ is the set of all words
from ${\cal G}^{(w_i)}(n+1)$ which contain some square as a suffix.
Denote by $\pi (i)$ the set of all quasi-ancestors of $w_i$.
Taking into account Proposition~\ref{utvizo}, it is easy to see that
\begin{equation}
|{\cal G}^{(w_i)}(n+1)|=\sum_{w\in\pi (i)} |{\cal F}_m^{(w)}(n)|.
\label{Lwin}
\end{equation}
We now estimate $|{\cal H}^{(w_i)}(n+1)|$. For any word~$w$ from
${\cal H}^{(w_i)}(n+1)$ we can find the minimal square which is
a suffix of~$w$. Denote the period of this square by $\lambda (w)$.
It is obvious that $\lfloor (m+1)/2\rfloor p$.
Note that in this case $j>m$. Let $w$ be an arbitrary word from
${\cal H}_j^{(w_i)}(n+1)$. Then for~$w$ we have $w[n-2j+2:n-j+1]=
w[n-j+2:n+1]$, i.e., $w[n-j-m+2:n-j+1]=w_i$ and $w$ is determined
uniquely by the prefix $w[1:n-j-m+1]$. Therefore, in this case
the inequality
\begin{equation}
|{\cal H}_j^{(w_i)}(n+1)|\le |{\cal F}_m^{(w_i)}(n-j+1)|.
\label{Ljwin1}
\end{equation}
holds. Using inequalities (\ref{Ljwin}) and~(\ref{Ljwin1})
in~(\ref{Lwin2}), we obtain
\begin{equation}
|{\cal H}^{(w_i)}(n+1)|\le A_p^{(w_i)}(n+1)+
B_p^{(w_i)}(n+1)
\label{Ljwin2}
\end{equation}
where
\begin{eqnarray*}
A_p^{(w_i)}(n+1)&=&\sum_{\lfloor (m+1)/2\rfloor 1$ and each $i=m, m+1,\ldots , n-1$ the inequality
$S^{\langle\mbox{sf}\rangle}_m(i+1)\ge \alpha S^{\langle\mbox{sf}\rangle}_m(i)$
be valid. Then for each $i=m, m+1,\ldots , n-1$ we have
$S^{\langle\mbox{sf}\rangle}_m(i)\le
S^{\langle\mbox{sf}\rangle}_m(n)/\alpha^{n-i}$. So relation~(\ref{sum_i_sx})
implies that
$$
\sum_{i=1}^s x_i\cdot A_p^{(w_i)}(n+1)\le \sum_{d=d_0}^q \rho_d\cdot
(S^{\langle\mbox{sf}\rangle}_m(n)/\alpha^d)=
{\cal P}_m^{(p,q)}(1/\alpha)\cdot S^{\langle\mbox{sf}\rangle}_m(n).
$$
In an analogous way relation~(\ref{sum_pjle}) implies that
\begin{eqnarray*}
\sum_{i=1}^s x_i B_p^{(w_i)}(n+1)&\le&\sum_{pS^{\langle\mbox{sf}\rangle}_m(n)\cdot
\left(r-{\cal P}_m^{(p,q)}(1/\alpha)-\frac{1}{\alpha^{p-1}(\alpha -1)}\right).
$$
Therefore, if $\alpha$ satisfy the inequality
$$
r-{\cal P}_m^{(p,q)}(1/\alpha)-\frac{1}{\alpha^{p-1}(\alpha -1)}
\ge\alpha,
$$
we obtain inductively that the inequality $S^{\langle\mbox{sf}\rangle}_m({n+1})
\ge \alpha S^{\langle\mbox{sf}\rangle}_m(n)$ holds for any~$n$. Thus,
in this case we have $S^{\langle\mbox{sf}\rangle}_m(n)=\Omega (\alpha^n)$.
Since, obviously, the order of growth of $S^{\langle\mbox{sf}\rangle}(n)$
is not less than $S^{\langle\mbox{sf}\rangle}_m(n)$, we then conclude that
$S^{\langle\mbox{sf}\rangle}(n)=\Omega (\alpha^n)$. Hence
$\gamma^{\langle\mbox{sf}\rangle}\ge\alpha$.
For obtaining a concrete lower bound on $\gamma^{\langle\mbox{sf}\rangle}$
we take the parameters $m=45$, $p=52$, $q=60$. Using computer computations,
we have obtained\footnote{In this paper the obtained numerical results are
given with precision of 6 decimal digits after the point.} that
$|{\cal F}''(45)|=277316$, the maximal in modulus eigenvalue~$r$ for
$\Delta_{45}$ is $1.302011$, and all components of the eigenvector
corresponding to~$r$ are positive. Further, we obtained that
$$
\begin{array}{c}
{\cal P}_{45}^{(52,60)}(z) = 3.759479\cdot z^{44} + 3.176743\cdot z^{45}
+ 6.048526\cdot z^{46} + \\
7.120005\cdot z^{48} + 14.679230\cdot z^{50} + 41.594270\cdot z^{52}
+ 37.431675\cdot z^{55} + \\
40.471892\cdot z^{56} + 32.780085\cdot z^{58}
+ 5.235193\cdot z^{59} + 275.705551\cdot z^{60}.
\end{array}
$$
Let $\alpha =1.30173$. It is immediately checked that
$$
r-{\cal P}_{45}^{(52,60)}(1/\alpha)-\frac{1}{\alpha^{51}(\alpha -1)}
\ge\alpha.
$$
Moreover, the inequalities $S^{\langle\mbox{sf}\rangle}_{45}(n+1)\ge
\alpha S^{\langle\mbox{sf}\rangle}_{45}(n)$ for each $n=45, 46,\ldots,
q+m-1=104$ are verified in the same inductive way as described above
with evident modifications following from the restriction $n2$. Analogously to the case of square-free words,
we introduce also the notions of descendant, ancestor, and closed word
for minimally repetitive words, and denote by ${\cal L}_m$ the set
of all words from $\Sigma^*_3$ which do not contain closed words from
${\cal F}(m)$ as factors. Denote also by ${\cal F}_m$ the set of all
minimally repetitive words from ${\cal L}_m$, by ${\cal F}'(m)$ the
set of all words~$w$ from ${\cal F}(m)$ such that $w[1]=0$ and $w[2]=1$,
and by ${\cal F}''(m)$ the set of all words from ${\cal F}'(m)$ which are
not closed. As in the case of square-free words, we introduce the notions
of quasi-descendant and quasi-ancestor, define for ${\cal F}''(m)$
the matrix $\Delta_m$ of size $s\times s$ where $s=|{\cal F}''(m)|$,
and compute the maximal in modulus eigenvalue~$r$ of this matrix.
If $r>1$ and all components of the eigenvector $\tilde x=(x_1;\ldots; x_s)$
corresponding to~$r$ are positive, then we denote by $\mu$ the ratio
$\max_i x_i/\min_i x_i$, and for $n\ge m$ consider
$S^{\langle\mbox{lf}\rangle}_m(n)=\sum_{i=1}^s x_i\cdot |{\cal F}_m^{(w_i)}(n)|$
where $w_i$ is $i$-th word of the set ${\cal F}''(m)$, $i=1,\ldots ,s$.
Analogously to equality~(\ref{Fwin}), for $i=1,\ldots ,s$ we have
\begin{equation}
|{\cal F}_m^{(w_i)}(n+1)|=|{\cal G}^{(w_i)}(n+1)|-|{\cal H}^{(w_i)}(n+1)|
\label{Fwin-1}
\end{equation}
where ${\cal G}^{(w_i)}(n+1)$ is the set of all words~$w$ from
${\cal L}_m^{(w_i)}(n+1)$ such that the words $w[1:n]$ and $w[n-m+1:n+1]$
are minimally repetitive, and ${\cal H}^{(w_i)}(n+1)$ is the set of all
words from ${\cal G}^{(w_i)}(n+1)$ which contain some prohibited
repetition as a suffix. Analogously to equality~(\ref{Lwin}), we
can obtain
$$
|{\cal G}^{(w_i)}(n+1)|=\sum_{w\in\pi (i)} |{\cal F}_m^{(w)}(n)|
$$
where $\pi (i)$ is the set of all quasi-ancestors of $w_i$.
For any word~$w$ from ${\cal H}^{(w_i)}(n+1)$ denote by~$\lambda (w)$
the minimal period of the shortest prohibited repetition which is
a suffix of~$w$. Then, analogously to equality~(\ref{Lwin2}),
$$
|{\cal H}^{(w_i)}(n+1)|=\sum_{\lfloor (4m+3)/7\rfloor \lfloor 7p/4\rfloor$.
Let $w$ be an arbitrary word from ${\cal H}_j^{(w_i)}(n+1)$
where $j\le p$. Then the suffix $w[n-\lfloor 7p/4\rfloor +1:n+1]$
is a prohibited repetition which contains neither closed words
from ${\cal F}(m)$ nor other prohibited repetitions as factors and
contains the word $w_i$ as a suffix. Let $v_1,\ldots,v_t$ be all
possible prohibited repetitions with minimal period~$j$ which satisfy
the given conditions. Denote by ${\cal H}_{j,k}^{(w_i)}(n+1)$ the set
of all words from ${\cal H}_j^{(w_i)}(n+1)$ which contain $v_k$
as a suffix, $k=1,\ldots, t$. Let $w\in {\cal H}_{j,k}^{(w_i)}(n+1)$.
Analogously to the case of square-free words, the symbol
$w[n-\lfloor 7j/4\rfloor ]$ is determined uniquely by $v_k$ as the
symbol from $\Sigma_3$ which is different from the two distinct
symbols $v_k[1]$ and $v_k[j]$. Denoting this symbol by $b_k$,
we conclude that the factor $w[n-\lfloor 7j/4\rfloor :n]$ is determined
uniquely as the word $b_k v_k[1:\lfloor 7j/4\rfloor ]$. Let this word
belong to ${\cal F}_m$. Then we denote by $u'_k$ the word from
${\cal F}'(m)$ which is isomorphic to the word $b_k v_k[1:m-1]$.
Analogously to inequality~(\ref{Ljwin}), one can obtain the inequality
$$
|{\cal H}_j^{(w_i)}(n+1)|\le\sum_{u\in U_j(w_i)}
|{\cal F}_m^{(u)}(n+m-\lfloor 7j/4\rfloor -1)|
$$
where $U_j(w_i)$ is the set of all words\footnote{Note that,
as in the case of square-free words, the same word can be counted
several times in $U_j(w_i)$.} $u'_k$. Thus,
\begin{equation}
|{\cal H}^{(w_i)}(n+1)|\le A_p^{(w_i)}(n+1)+
\sum_{pp$, i.e., $j\ge 4m/3$. Note that the sets
${\cal H}_j^{(w_i)}(n+1)$ are non-overlapping. So we have
the obvious inequality
\begin{equation}
\sum_{i=1}^s x_i\cdot |{\cal H}_j^{(w_i)}(n+1)|\le
|{\cal M}_j|\cdot \max_{i=1,\ldots,s} x_i
\label{sumHj}
\end{equation}
where ${\cal M}_j=\bigcup_{i=1}^s {\cal H}_j^{(w_i)}(n+1)$.
Note also that any word $w$ from ${\cal M}_j$ is determined
uniquely by the prefix $w[1:n-\lfloor 3j/4\rfloor ]$ and
satisfies the conditions $w[n+2-m-j]=w[n+2-m]=0$ and
$w[n+3-m-j]=w[n+3-m]=1$. Thus $|{\cal M}_j|\le |{\cal M}'_j|$
where ${\cal M}'_j$ is the set of all words $w$ from
${\cal F}_m(n-\lfloor 3j/4\rfloor )$ such that $w[n+2-m-j]=0$
and $w[n+3-m-j]=1$. Consider also the set ${\cal M}''_j$ of all
words $w$ from ${\cal F}_m(n-\lfloor 3j/4\rfloor )$ such that
$w[n+1-\lfloor 3j/4\rfloor -m]=0$ and $w[n+2-\lfloor 3j/4\rfloor -m]=1$.
There is an evident bijection between the sets ${\cal M}'_j$ and
${\cal M}''_j$, so $|{\cal M}'_j|=|{\cal M}''_j|$. Note also that
the set ${\cal M}''_j$ is the union of the non-overlapping sets
${\cal H}_j^{(w_i)}(n-\lfloor 3j/4\rfloor )$ for $i=1,\ldots, s$,
i.e.,
$$
|{\cal M}''_j|\le \sum_{i=1}^s |{\cal H}_j^{(w_i)}(n-\lfloor 3j/4\rfloor )|
\le S^{\langle\mbox{lf}\rangle}_m(n-\lfloor 3j/4\rfloor )/
(\min_{i=1,\ldots,s} x_i).
$$
Therefore, it follows from~(\ref{sumHj}) that
$$
\sum_{i=1}^s x_i\cdot |{\cal H}_j^{(w_i)}(n+1)|\le
|{\cal M}'_j|\cdot \max_{i=1,\ldots,s} x_i =
|{\cal M}''_j|\cdot \max_{i=1,\ldots,s} x_i \le
\mu\cdot S^{\langle\mbox{lf}\rangle}_m(n-\lfloor 3j/4\rfloor ).
$$
Thus,
\begin{equation}
\sum_{p1$ and each $i=m, m+1,\ldots , n-1$ the inequality
$S^{\langle\mbox{lf}\rangle}_m(i+1)\ge \alpha S^{\langle\mbox{lf}\rangle}_m(i)$
be valid, i.e., $S^{\langle\mbox{lf}\rangle}_m(i)\le
S^{\langle\mbox{lf}\rangle}_m(n)/\alpha^{n-i}$. Then
\begin{equation}
\sum_{i=1}^s x_i\cdot A_p^{(w_i)}(n+1)\le \sum_{d=d_0}^q \rho_d\cdot
(S^{\langle\mbox{lf}\rangle}_m(n)/\alpha^d)=
{\cal P}_m^{(p,q)}(1/\alpha)\cdot S^{\langle\mbox{lf}\rangle}_m(n)
\label{sumA_p}
\end{equation}
where ${\cal P}_m^{(p,q)}(z)=\sum_{d=d_0}^q \rho_d\cdot z^d$.
Moreover, it follows from~(\ref{sumplej}) that
$$
\begin{array}{c}
\displaystyle
\sum_{p&S^{\langle\mbox{lf}\rangle}_m(n)
\cdot\Biggl(r-{\cal P}_m^{(p,q)}(1/\alpha)-\\
&&\mu\cdot\Bigl(1/\bigl(\alpha^{\lfloor (3p-1)/4\rfloor} (\alpha -1)\bigr)
+1/\bigl(\alpha^{3\lfloor p/4\rfloor}(\alpha^3-1)\bigr)\Bigr)\Biggr).
\end{eqnarray*}
Therefore, if
$$
r-{\cal P}_m^{(p,q)}(1/\alpha)-\mu\cdot\Bigl(
1/\bigl(\alpha^{\lfloor (3p-1)/4\rfloor} (\alpha -1)\bigr)+
1/\bigl(\alpha^{3\lfloor p/4\rfloor}(\alpha^3-1)\bigr)\Bigr)
\ge\alpha
$$
then $S^{\langle\mbox{lf}\rangle}_m(n+1)\ge
\alpha S^{\langle\mbox{lf}\rangle}_m(n)$ for any~$n$,
i.e., $S^{\langle\mbox{lf}\rangle}_m(n)=\Omega (\alpha^n)$.
Since the order of growth of $S^{\langle\mbox{lf}\rangle}(n)$
is not less than $S^{\langle\mbox{lf}\rangle}_m(n)$, in this case
we have $S^{\langle\mbox{lf}\rangle}(n)=\Omega (\alpha^n)$,
i.e., $\gamma^{\langle\mbox{lf}\rangle}\ge\alpha$.
Using computer computations with the parameters $m=42$, $p=72$, $q=85$,
we obtained that $|{\cal F}''(42)|=36141$, $r=1.247500$, all components
of the eigenvector corresponding to~$r$ were positive, and
${\cal P}_{42}^{(72,85)}(z)$ was
$$
\begin{array}{l}
1.976268\cdot z^{42} + 1.148062\cdot z^{44}
+ 3.519576\cdot z^{45} + 1.741046\cdot z^{47} + \\
9.687624\cdot z^{49} + 0.126312\cdot z^{50}+ 31.479339\cdot z^{52} +
12.284335\cdot z^{53} + \\
21.010557\cdot z^{54} + 24.183001\cdot z^{56} + 96.529327\cdot z^{61} +
129.216325\cdot z^{64} + \\
256.213310\cdot z^{66} + 14.826731\cdot z^{67} + 64.163103\cdot z^{68} +
6.862805\cdot z^{69} + \\
84.819931\cdot z^{70} + 2.337610\cdot z^{72} + 175.026144\cdot z^{73} +
41.068102\cdot z^{74} + \\
335.714818\cdot z^{75} + 341.576384\cdot z^{78} + 329.970329\cdot z^{80} +
693.282157\cdot z^{81} + \\
763.104210\cdot z^{82} + 303.272754\cdot z^{83} +
583.157071\cdot z^{84} + 10510.070498\cdot z^{85}.
\end{array}
$$
Let $\alpha =1.245$. It is immediately checked that
$$
r-{\cal P}_{42}^{(72,85)}(1/\alpha)-\frac{1}{\alpha^{53}(\alpha-1)}-
\frac{1}{\alpha^{54}(\alpha^3-1)}\ge\alpha.
$$
Moreover, we estimate $S^{\langle\mbox{lf}\rangle}_{42}(n+1)\ge
\alpha S^{\langle\mbox{lf}\rangle}_{42}(n)$ for each $n=42, 43,\ldots ,
q+m-1=126$ in the same inductive way with evident modifications
following from the restriction $n2$ we can introduce the notions of descendant,
ancestor, and closed word for cube-free words. Denote also by ${\cal L}_m$
the set of all words from $\Sigma^*_2$ which do not contain closed words
from ${\cal F}(m)$ as factors, and by ${\cal F}_m$ the set of all cube-free
words from ${\cal L}_m$. By ${\cal F}'(m)$ we denote the set of all words~$w$
from ${\cal F}(m)$ such that $w[1]=0$. Note that for any word~$w$ from
${\cal F}(m)$ there exists a single word from ${\cal F}'(m)$ which is
isomorphic to~$w$. By ${\cal F}''(m)$ we denote the set of all words from
${\cal F}'(m)$ which are not closed. We introduce also the notions of
quasi-descendant and quasi-ancestor, define for ${\cal F}''(m)$
the matrix $\Delta_m$ of size $s\times s$ where $s=|{\cal F}''(m)|$,
and compute the maximal in modulus eigenvalue~$r$ of this matrix.
If $r>1$ and all components of the eigenvector $\tilde x=(x_1;\ldots; x_s)$
corresponding to~$r$ are positive, then for $n\ge m$ we consider
$S^{\langle\mbox{cf}\rangle}_m(n)=\sum_{i=1}^s x_i\cdot |{\cal F}_m^{(w_i)}(n)|$
where $w_i$ is $i$-th word of the set ${\cal F}''(m)$, $i=1,\ldots ,s$.
As in the case of square-free words, for $i=1,\ldots ,s$ we have
\begin{equation}
|{\cal F}_m^{(w_i)}(n+1)|=|{\cal G}^{(w_i)}(n+1)|-|{\cal H}^{(w_i)}(n+1)|
\label{Fwin-3}
\end{equation}
where ${\cal G}^{(w_i)}(n+1)$ is the set of all words~$w$ from
${\cal L}_m^{(w_i)}(n+1)$ such that the words $w[1:n]$ and $w[n-m+1:n+1]$
are cube-free, and ${\cal H}^{(w_i)}(n+1)$ is the set of all words
from ${\cal G}^{(w_i)}(n+1)$ which contain some cube as a suffix.
Analogously to equality~(\ref{Lwin}), we obtain
$$
|{\cal G}^{(w_i)}(n+1)|=\sum_{w\in\pi (i)} |{\cal F}_m^{(w)}(n)|
$$
where $\pi (i)$ is the set of all quasi-ancestors of $w_i$.
For any word~$w$ from ${\cal H}^{(w_i)}(n+1)$ denote by~$\lambda (w)$
the period of the minimal cube which is a suffix of~$w$. Then,
analogously to equality~(\ref{Lwin2}),
\begin{equation}
|{\cal H}^{(w_i)}(n+1)|=\sum_{\lfloor (m+1)/3\rfloor p$, analogously to
inequality~(\ref{Ljwin1}), we have
\begin{equation}
|{\cal H}_j^{(w_i)}(n+1)|\le |{\cal F}_m^{(w_i)}(n-2j+1)|.
\label{Ljwin-1}
\end{equation}
Thus, from (\ref{Lwin-2}), (\ref{Ljwin0}) and~(\ref{Ljwin-1})
we obtain that
\begin{equation}
|{\cal H}^{(w_i)}(n+1)|\le A_p^{(w_i)}(n+1)+
B_p^{(w_i)}(n+1)
\label{Ljwin3}
\end{equation}
where
\begin{eqnarray*}
A_p^{(w_i)}(n+1)&=&\sum_{\lfloor (m+1)/3\rfloor 1$ and each $i=m, m+1,\ldots , n-1$ the inequalities
$S^{\langle\mbox{cf}\rangle}_m(i+1)\ge \alpha S^{\langle\mbox{cf}\rangle}_m(i)$
be valid. Then
$$
\sum_{i=1}^s x_i\cdot A_p^{(w_i)}(n+1)\le \sum_{d=d_0}^q \rho_d\cdot
(S^{\langle\mbox{cf}\rangle}_m(n)/\alpha^d)={\cal P}_m^{(p,q)}(1/\alpha)\cdot
S^{\langle\mbox{cf}\rangle}_m(n),
$$
where ${\cal P}_m^{(p,q)}(z)=\sum_{d=d_0}^q \rho_d\cdot z^d$.
Moreover, it follows from~(\ref{sum_pjl1}) that
$$
\sum_{i=1}^s x_i\cdot B_p^{(w_i)}(n+1)\le \sum_{p&S^{\langle\mbox{cf}\rangle}_m(n)\cdot\left(r-{\cal P}_m^{(p,q)}(1/\alpha)-
\frac{1}{\alpha^{2p-1}(\alpha^2-1)}\right).
\end{eqnarray*}
Therefore, if
$$
r-{\cal P}_m^{(p,q)}(1/\alpha)-\frac{1}{\alpha^{2p-1}(\alpha^2-1)}
\ge\alpha,
$$
then $S^{\langle\mbox{cf}\rangle}_m({n+1})\ge \alpha
S^{\langle\mbox{cf}\rangle}_m(n)$ for any~$n$, i.~e.
$S^{\langle\mbox{cf}\rangle}_m(n)=\Omega (\alpha^n)$.
Since the order of growth of $S^{\langle\mbox{cf}\rangle}(n)$
is not less than $S^{\langle\mbox{cf}\rangle}_m(n)$, we
obtain in this case that $S^{\langle\mbox{cf}\rangle}(n)=
\Omega (\alpha^n)$, i.~e. $\gamma^{\langle\mbox{cf}\rangle}\ge\alpha$.
Using computer computations with the parameters $m=35$, $p=35$, $q=70$,
we obtained that $|{\cal F}''(35)|=732274$, $r=1.457599$, all components
of the eigenvector corresponding to~$r$ were positive, and
${\cal P}_{35}^{(35,70)}(z)$ was
$$
\begin{array}{l}
0.890340\cdot z^{35} + 1.398382\cdot z^{37}
+ 1.096456\cdot z^{38} + 30.292784\cdot z^{40} + \\
2.533687\cdot z^{41} + 1.296919\cdot z^{42} + 28.893958\cdot z^{43} +
22.780262\cdot z^{44} + \\
10.699704\cdot z^{45} + 64.314464\cdot z^{47} + 92.853910\cdot z^{49} +
91.743094\cdot z^{50} + \\
67.688387\cdot z^{51} + 48.613345\cdot z^{52} + 68.285930\cdot z^{53} +
113.239316\cdot z^{54} + \\
144.612325\cdot z^{56} + 346.136318\cdot z^{58} + 173.468149\cdot z^{59} +
465.000388\cdot z^{60} + \\
134.993653\cdot z^{61} + 224.831969\cdot z^{62} + 585.928351\cdot z^{63} +
355.591901\cdot z^{65} + \\
1335.518621\cdot z^{67} + 343.074473\cdot z^{68} + 2202.468159\cdot z^{69} +
11098.126369\cdot z^{70}.
\end{array}
$$
It is immediately checked that for $\alpha =1.457567$ the inequality
$$
r-{\cal P}_{35}^{(35,70)}(1/\alpha)-\frac{1}{\alpha^{69}(\alpha^2-1)}
\ge\alpha
$$ is valid. Moreover, the inequalities $S^{\langle\mbox{cf}\rangle}_{35}(n+1)
\ge \alpha S^{\langle\mbox{cf}\rangle}_{35}(n)$ for $n=35, 36,\ldots ,q+m-1=104$
are also verified in the same inductive way with evident modifications
following from the restriction $n